Question

我在目录中有几个项目，我想编写一个Makefile来构建所有这些项目的或子集。每个项目都位于一个以自己命名的文件夹中，该文件夹包含一个Makefile。

如何执行这些Makefile？

这是我尝试过的：

# Define the project names
PROJECT_NAMES := \
    Project_1 \
    Project_2 \
    Project_3

# Define default behaviour
default: all

# Rule to build all projects
all:
    $(foreach project, $(PROJECT_NAMES), $(CURRENT_PROJECT))

# Rule to build single project
.PHONY $(CURRENT_PROJECT)
$(CURRENT_PROJECT):
    $(MAKE) -C $(CURRENT_PROJECT) make

我认为这个问题可能适用于类似问题，但未得到回答： How to make a Makefile call another Makefile rules?

Answer 1

应该使用先决条件/依赖性来代替传递参数。这是我的解决方法：

# Define the project names
PROJECT_NAMES := \
    Project_1 \
    Project_2 \
    Project_3

# Define default behaviour
default: all

# Rule to build all projects now depends on building individual projects
all: $(foreach project, $(PROJECT_NAMES), $(project)_build)

# Rule to build a single project
.PHONY: Project_%
Project_%:
    @echo "****** Building $(subst _build,,$@) ******"
    @$(MAKE) -C $(subst _build,,$@) all

在不带参数的情况下调用所有项目。只需将项目名称传递到makefile中即可构建单个项目的任何组合，例如：

make Project_1 Project_3

如何将变量传递给Makefile中的规则？

1 个答案: